Tuesday, September 15, 2015

I've been having trouble getting the Widmark formula and factor to work out properly for Widmark's 1914 data. One can use the intercept of the vertical concentration axis and slope to determine the line which represents the Widmark formula and which in this case is B' = B0 - IBt. The Widmark factor, r, obtained from the B0 = A/(Mr) should be approximately 0.70 but the calculated value is 0.32 instead which is too low. Widmark showed that the concentration of alcohol in urine was approximately the same as that in the blood. So it would appear that the alcohol concentration values for the data are too high.

Thursday, September 10, 2015

Widmark expresses the blood alcohol content as per mille, ‰, but this term can at first seem a little vague. By studying Table 1 in Neymark and Widmark, Kinetik des Aethylalkoholumsatzes (1935), we can assign a more meaningful unit to measure the BAC.

It appears from the data in the table that ‰ is equivalent to 1 gm/1000ml but since the density of water is 1 kg/liter this could also mean 1 gm/kg or 1 gm/1000gm which is unitless. Since by concentration we usually mean quantity per unit volume it seems more reasonable to use 1 gm/1000ml. When we do this the unit for the Widmark factor is liter/kg which is what we used before. Even though the factor here was obtained by experiments on dogs, we still get r = 0.7 liter/kg approximately.

Another expression, per os, used in Widmark, Konzentration des Alkohols (1914), means taken by mouth so it appears he drank the alcohol in a number of different concentrations but obtained similar results for the BAC vs time. In the experiments on the dogs the alcohol was administered by intravenous injection directly into the blood.

Monday, September 7, 2015

I found the degenerate solution for the diffusion model used in the last posting and tried doing a least squares fit again. I used the trick guessing at where the initial zero, t0, and the final zero, tR, would be as well as "relative likelihood" statistical weights for the sums involved. There was only one parameter, α, to search for since IB was found using the least squares fit. The result was closer to what one would expect from statistical errors in the observations coming within two standard deviations of the data points.

The improvement in the fit came from using the statistical weights which were inversely proportional to the standard deviations of the data points.

Saturday, September 5, 2015

Using a few tricks I was able to get a fit for the blood alcohol content data published by Widmark in 1914 for a 4-compartment model. The flow was modeled by the following analogous electrical circuit with I0 representing the removal rate.

There wasn't very much data to work with and at first my fits looked a little strange but I decided to estimate the time at which the BAC was zero and adjusted those for the best least squares fit of the general solution of the assumed model. One criterion for the fit was that the downward slope be approximately linear. Another problem was that the best fit again appears to be degenerate suggesting that the two time constants are equal.

The peak value still appears to be a little off but the fit would probably benefit from more accurate data over a longer span of time. Only positive values for the BAC fit make any physical sense. The fitted curve can act a little weird to the left and right of the region shown but the assumed removal rate isn't valid in these regions either.

Wednesday, September 2, 2015

I used an electrical RC circuit to model the flow of alcohol from an initial compartment to two others with no removal of the alcohol from the system to simplify the problem. Alcohol diffusing from the digestive tract into the blood and from there into body tissues can show a slightly higher peak than the 3-compartment model. The circuit used, component values and some of its initial conditions are as follows.

One can compute successive values for the charges on the capacitors which are analogous to the amount of alcohol in each compartment using the simultaneous equations or attempt to solve the system of equations analytically. The analytical equations are easier to compute but can be rather difficult to find. The solutions for the charges on the capacitors are shown in the figure below. I included the analytical equation for the charge, qB, corresponding to the blood alcohol content. Notice that the digestive tract and blood can approach equilibrium before the equilibrium is reached for the tissues accounting for the slightly raised peak.

Since qB and qT are initially zero the simultaneous equations tell us that the initial rate of change of qT is also zero and its curve is slow to change at first. The characteristic equation for the circuit is cubic but fortunately the constant term is missing so one only has to solve a quadratic equation to get the values of α1 and α2.